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Consider a continuous function $f:\mathbb{R}\rightarrow[0,\infty)$ that does not tend to zero as its argument tends to infinity. Formally, there is some $\varepsilon>0$ such that there does not exist a $T\in\mathbb{R}$ for which

$$t\geq T\Rightarrow f(t)\leq \varepsilon.$$

Is it true that there exists some $\alpha>0$ such that for any $t_0,t_1\in\mathbb{R}$

$$\int_{t_0}^{t_1}f(s)ds\geq\alpha(t_1-t_0).$$

I've been trying to think of a counterexample, but continuity is making it difficult.

Thanks.

EDIT: Sorry, the question was meant to be:

"Is it true that there exists some $\alpha>0$ such that for any $t_0,t_1\in\mathbb{R}$, with $t_1\geq t_0$, one can always find a sufficiently large $t_2\geq t_1$ so that

$$\int_{t_0}^{t_2}f(s)ds\geq\alpha(t_2-t_0)."$$

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Drawing a picture of what's going on may help here. –  tharris Apr 22 '13 at 15:48
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2 Answers

up vote 1 down vote accepted

You can just make the integral zero for some choices of $t_i$ by making $f$ itself zero at such places. Then you only need to add lots of spikes to make it not converge to 0.

Edit: enter image description here

Second edit for edited question: Just place the spikes so that they are identical, area 1, and centred at $x=n^2$ for each $n\in\mathbb Z$. For bad choices of $t_0$, integral is effectively bounded by something like $t_1^{1/2}-t_0^{1/2}$ so the statement is whether there is $\alpha>0$ such that you can find $t_1\ge t_0$ $t_1^{1/2}-t_0^{1/2} \ge \alpha (t_1-t_0) \iff \alpha^{-1} \ge t_1^{1/2} + t_0^{1/2}$ regardless of how big $t_0$ is. This clearly does not hold.

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Agreed, however I'm having trouble adding the spikes without breaking continuity. If you had a fixed $\alpha$ this would be easy, but its the "for any" that is causing me trouble. Can think of an actual counterexample? –  jkn Apr 22 '13 at 15:57
    
@jkn try $f_n$ with support in $[n,n+1/n^\beta]$ for a suitably chosen $\beta \geq 1$, and then define $f = \sum_n f_n$ (which happens to equal $\sup_n f_n$ which may or may not be useful to you). –  kahen Apr 22 '13 at 16:09
    
@Sharkos, aha! Thanks. –  jkn Apr 22 '13 at 16:21
    
@jkn - spikes as shown are obviously continuous. If you make $f([a,b]) = 0$ then $\int_{t_0}^{t_1} f = 0$ for all $t_i\in [a,b]$ so you don't care about $\alpha$. –  Sharkos Apr 22 '13 at 16:21
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I'm not revising this answer any more times, this is getting silly... –  Sharkos Apr 22 '13 at 16:45
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It is false. If it were true then $$ \lim_{t\to\infty}\int_{t_0}^tf(s)\,ds=\infty, $$ which is not the case if $f$ is integrable (in the Lebesgue sense, or as an improper integral if you are talking about Riemann integrals) on $\mathbb{R}$.

Counterexample

Let $$ T(x)=\begin{cases} 1-|x| & \text{if }|x|\le1,\\0 & \text{if }|x|>1, \end{cases} $$ and define $$ f(x)=\sum_{n=1}^\infty T(2^n(x-n)). $$ $T(2^n(x-n))$ is supported on $[n-2^{-n},n+2^{-n}]$ and its integral is $2^{-n}$, so that $$ \int_{-\infty}^{+\infty}f(x)\,dx=\sum_{n=1}^\infty2^{-n}=1\text{ and }f(n)=1. $$

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Can't the unsigned Lebesgue integral take values in $[0,+\infty]$? What do you mean by integrable here? –  jkn Apr 22 '13 at 16:00
    
Presumably that it is in $\mathcal L^1$? –  kahen Apr 22 '13 at 16:04
    
(I'm not assuming $f\in\mathcal{L}^1$, all I'm assuming is that $f\in\mathcal{C}$). –  jkn Apr 22 '13 at 16:09
    
If Lebesgue integral: $f$ is measurable and $\int_{-\infty}^{+\infty}f(x)\,dx<\infty$. If Riemann integral: $\int_{-\infty}^{+\infty}f(x)\,dx$ exists as an improper integral- –  Julián Aguirre Apr 22 '13 at 16:12
    
Pretty sure the examples you've given tend to 0 as $x\to\infty$. –  Sharkos Apr 22 '13 at 16:16
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